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Functions, operator => eigenfunction, eigenvalue 
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#1
Sep2407, 05:57 AM

P: 12

1. The problem statement, all variables and given/known data
Show, that functions f1 = A*sin([tex]\theta[/tex])exp[i[tex]\phi[/tex]] and f2 = B(3cos[tex]^{2}[/tex]([tex]\theta[/tex])  1) A,B  constants are eigenfunctions of an operator and find eigenvalues 3. The attempt at a solution This is what i got for the first function: The next step is to solve left part of the equation, and than compare it to the right part. The question arises, how to solve that equation? I tried simplifying left part of an equation in mathcad, and I got Next question from that part, is if I am doing it right, how to compare those parts, and answer a question  weather this function is an eigenfunction of an operator? Thank You in advance, and I am constantly near computer and waiting for suggestions. 


#2
Sep2407, 09:24 AM

P: 777

Your eigen operator has partial derivatives wrt to [itex]\theta[/itex] and [itex]\phi[/itex]. When you operate it on your given eigen function, you should get back your original function multiplied by a scaling factor which is your eigenvalue.



#3
Sep2407, 11:25 AM

P: 12

Yes, Thank You, but how do I calculate that? May I ask for instructions on how to calculate that left part of an equation? Is the result I got is correct?



#4
Sep2407, 11:32 AM

Sci Advisor
HW Helper
P: 2,887

Functions, operator => eigenfunction, eigenvalue
all you have to do is to apply the derivatives 


#5
Sep2407, 11:34 AM

Sci Advisor
HW Helper
P: 2,887

[tex] \frac{1}{sin \theta} \frac{\partial}{\partial \theta} ( sin \theta ~\frac{\partial}{ \partial \theta} (A sin \theta} e^{i \phi}}) ) [/tex] 


#6
Sep2407, 12:37 PM

P: 12

I will give it a try right know. Thank You.



#7
Sep2407, 01:58 PM

P: 12



#8
Sep2407, 09:27 PM

P: 69

The part in brackets is an "operator"... every incomplete differentiation sign (ie, without anything to differentiate) operates on whatever is "multiplied" to it.
For example, the d/d(theta) is your image also operates on A*sin(theta)*exp(i*phi), because when you open the bracket, it gets to differentiate that term. Eg. (the d's are partial) [d/dx + d/dy]*x*y^2 = y^2 + 2*y*x 


#9
Sep2507, 08:21 AM

P: 777




#10
Sep2507, 11:04 AM

P: 12

Thank You Reshma! I already managed to solve it correctly, and find eigenvalues as well.
For the case with the first function : a = 2h^2, and for the case with second function b = 6h^2. The problem was initially in understanding how to apply operator properly. Once example have been given, and operator properties reanalyzed  problem got solved in like 10 minutes (with putting it all on a paper as well). Thank You everybody who took part in this one! 


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